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n200080 [17]
3 years ago
9

Can somebody plz help answer these false and truth questions correctly (only if u know for sure) thx :3

Chemistry
2 answers:
AnnyKZ [126]3 years ago
8 0

Answer:

1. false

2. true

3.true

4.true

Explanation:

fenix001 [56]3 years ago
5 0

Answer:

1. True

2. True

3. True

4. True

5. True

HOPE THIS HELPS

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Hey!!! Chemistry Question about MOLES... Please help...
yulyashka [42]

Answer:

= 13.0 moles O2

Explanation:

1] Given the equation: 2C8H18 + 25 O2 ----> 16CO2 + 18H2O

a. How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide?

8.33 moles CO2 X

25mol O2

16mol CO2

= 13.0 moles O2

6 0
2 years ago
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What mass of Iron (III) acetate will be produced when 1.50 grams of iron reacts in acetic acid?
pantera1 [17]

Answer:

2Fe + 6HC2H3O2 → 2Fe(C2H3O2)3 + 3H2

Explanation:

There you go

4 0
3 years ago
You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0
3 years ago
What reactions are responsible for the glow and heat from the sun? nuclear fission nuclear fusion chemical reactions atomic disi
Mamont248 [21]
Nuclear Fusion is the answer to the question who posted.

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3 years ago
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Lead (II)oxide+calcium metal=​
AfilCa [17]
I didn’t understand
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